A 100-m-long high-voltage cable is suspended between two towers.The mass of the 100-m cable is 150 kg.If the tension in the cable is 30 000 N,what is the lowest frequency at which this cable can oscillate?
A)0.71 Hz
B)1.0 Hz
C)1.4 Hz
D)2.0 Hz

Question

Quizplus · Accepted Answer

The lowest frequency at which a cable can oscillate is determined by its fundamental frequency. The formula for the fundamental frequency of a vibrating string is f = (1/2L) * sqrt(T/μ), where L is the length, T is the tension, and μ is the mass per unit length. Here, L = 100 m, T = 30,000 N, and μ = 150 kg / 100 m = 1.5 kg/m. Plugging these values into the formula gives f = (1/200) * sqrt(30,000/1.5) = 0.71 Hz.

A 100-M-Long High-Voltage Cable Is Suspended Between Two Towers